According to the Kirchhoff law, for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.
The, the emissivity coefficient \epsilon is equal to the absorption coefficient \alpha, at least for a given wavelangth.
For glass, the emissivity is \epsilon= 0.9, while the value of \alpha seems to be variable between, approx, 0.2 and 0.5.
Does the Kirchhoff law hold for transparent materials? Or, on the contrary, one has \epsilon=\alpha+\tau (where \tau is the transmission coefficient)
...Or maybe the mentioned difference is only due to the different wavelength?
Quantitatively, emissivity is the ratio of the thermal radiation from a surface to the radiation from an ideal black surface at the same temperature as given by the Stefan–Boltzmann law.
this paper can help you
Article Spectral Emissivity and Absorption Coefficient of Silica Gla...
Dear Laura,
Due to reciprocity, the emittance is equal to the absorptance at any wavelengths for common materials, including transparent materials. Generally, the absorptance (emittance) is obtained by measuring the transmittance and reflectance and using the equation A= 1-T-R. For most of the glasses (macroscale thickness), the visible absorptance is very small but the mid-IR absorptance is high (0.8-0.9) especially in the 8-13 μm. Absorptance (emittance) of materials depends on the absorption coefficient that is derived from the penetration depth of light. https://en.wikipedia.org/wiki/Attenuation_coefficient
Gardon, Robert. "The emissivity of transparent materials." Journal of the American Ceramic Society 39, no. 8 (1956): 278-285.
Quantitatively, emissivity is the ratio of the thermal radiation from a surface to the radiation from an ideal black surface at the same temperature as given by the Stefan–Boltzmann law.
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